A body acquires charge $8.0 \times 10^{-12}$ C. The mass of the body: (A) increases by $4.5 \times 10^{-7}$ kg (B) decreases by $1.0 \times 10^{-6}$ kg (C) decreases by $4.55 \times 10^{-23}$ kg (D) increases by $9.1 \times 10^{-23}$ kg

> [!TLDR] > When a body gains positive charge, it loses electrons; the mass change equals the number of electrons lost times the electron mass. For $8.0 \times 10^{-12}$ C, the body **decreases** by $4.55 \times 10^{-23}$ kg. ## Why charging changes mass Charging a body means adding or removing electrons. Since electrons have mass ($m_e = 9.1 \times 10^{-31}$ kg), any change in the number of electrons changes the body's total mass. The sign of the charge tells us what happened: - **Positive charge**: electrons were removed → mass decreases - **Negative charge**: electrons were added → mass increases The magnitude of charge tells us *how many* electrons moved, since each electron carries charge $e = 1.6 \times 10^{-19}$ C. ## Finding the mass change 1. **Determine the number of electrons involved** The charge acquired is $Q = 8.0 \times 10^{-12}$ C (positive). The number of electrons that must have been removed is: $$n = \frac{Q}{e} = \frac{8.0 \times 10^{-12}}{1.6 \times 10^{-19}}$$ $$n = 5.0 \times 10^{7} \text{ electrons}$$ 2. **Calculate the total mass of these electrons** Each electron has mass $m_e = 9.1 \times 10^{-31}$ kg, so the total mass lost is: $$\Delta m = n \times m_e = 5.0 \times 10^{7} \times 9.1 \times 10^{-31}$$ $$\Delta m = 45.5 \times 10^{-24} = 4.55 \times 10^{-23} \text{ kg}$$ 3. **Determine the direction of change** Since the body acquired *positive* charge, electrons left the body. Therefore, the mass **decreases** by $4.55 \times 10^{-23}$ kg. > [!WARNING] > A common mistake is to forget that positive charge means electron *loss*, not gain. The body doesn't gain positive particles; it loses negative ones. > [!TIP] > Quick check: the mass change in charging problems is always tiny (order $10^{-23}$ kg or smaller for typical lab charges) because electron mass is so small. Options with larger mass changes are usually wrong. > [!ANSWER] > The correct option is **(C)**: the mass decreases by $4.5

A body acquires charge 8.0 × 10⁻¹² C. The mass of the body: (A) increases by 4.5 × 10⁻⁷ kg (B) decreases by 1.

Question

8.0 \times 10^{-12}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Quantization of Charge

Quantization of Charge

Imagine you are at a water fountain. You can fill your bottle with any amount of water — a little, a lot, or anything in between. Water is continuous. Now imagine instead that you are buying marbles. You can only buy marbles in whole numbers: 1 marble, 2 marbles, 15 marbles. You cannot buy half a marble or 2.7 marbles. Electric charge behaves like marbles, not like water.

That is the core intuition: charge comes in discrete packets. You cannot have an arbitrary amount of charge. You can only have whole-number multiples of a smallest possible chunk.

The smallest chunk: the elementary charge

That smallest chunk is called the elementary charge, denoted by the symbol $e$ . Its value is:

e = 1.602 \times 10^{-19} \text{ coulombs}

This is the magnitude of charge carried by a single proton (positive) or a single electron (negative). A proton has charge $+e$ , an electron has charge $-e$ .

Important

Every charged object in the universe — from a rubbed balloon to a lightning bolt — carries a total charge that is an integer multiple of $e$ . No exception has ever been observed among free, isolated charges. (Quarks carry fractional charges of $\pm e/3, \pm 2e/3$ but are always confined inside composite particles such as protons and neutrons, whose own net charge is still an integer multiple of $e$ .)

The precise statement

If $q$ is the total charge on any object, then:

q = n e

where $n$ is an integer ( $n = 0, \pm 1, \pm 2, \pm 3, \dots$ ).

The sign of $n$ tells you whether the charge is positive or negative. The magnitude $|n|$ tells you how many elementary charges are present (in excess or deficit).

q = n e, \quad n \in \mathbb{Z}

Why this matters

This is not a mathematical trick. It is a fundamental law of nature. It means:

You cannot have a charge of $0.5e$ or $1.7e$ .
If you measure the charge on any object, you will always find it to be $0$ , $\pm e$ , $\pm 2e$ , $\pm 3e$ , and so on.
All charge transfer — rubbing, conduction, induction — happens by moving whole electrons or protons. You cannot transfer a fraction of an electron.

Watch out

A common mistake is to think that because charge values like $3.2 \times 10^{-19} \text{ C}$ look like decimals, they are not multiples of $e$ . But $3.2 \times 10^{-19} \text{ C}$ is exactly $2e$ (since $2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}$ ). Always check by dividing by $e$ — the result must be an integer.

A concrete example

A glass rod rubbed with silk acquires a charge of $+4.8 \times 10^{-19} \text{ C}$ . How many electrons were transferred?

Divide the total charge by $e$ :

n = \frac{+4.8 \times 10^{-19}}{1.6 \times 10^{-19}} = +3

So the rod lost exactly 3 electrons. It could not have lost 2.5 or 3.7 electrons. The charge is $+3e$ .

Why this is called "quantization"

In physics, a quantity that can only take discrete, separated values is said to be quantized. Charge is quantized. Energy, in many contexts, is also quantized (think of atomic energy levels). The word comes from the Latin quantus — "how much" — and it signals that nature, at a fundamental level, is not smooth but grainy.

Note

In everyday life, charges are huge (coulombs contain about $6 \times 10^{18}$ elementary charges), so the graininess is invisible — just like sand looks smooth from a distance. But at the microscopic scale, the discrete nature of charge is absolute.

Final takeaway

Quantization of charge means: charge is not a continuous fluid. It comes in indivisible packets of size $e$ . Every charge $q$ satisfies $q = ne$ , where $n$ is an integer. This is one of the most fundamental facts about electricity — and it follows directly from the existence of electrons and protons as discrete particles.

Quantization of charge is one of the earliest concepts introduced in the NCERT Class 12 Physics Electrostatics chapter, and 'quantization of charge formula q = ne' or 'quantization of charge important questions class 12 physics' are frequent exam-prep searches. This fundamental fact about electrons and protons is also a reliable one-mark or assertion-reasoning question in board exams and JEE Main.

When a body gains positive charge, it loses electrons; the mass change equals the number of electrons lost times the electron mass. For $8.0 \times 10^{-12}$ C, the body decreases by $4.55 \times 10^{-23}$ kg.

Why charging changes mass

Charging a body means adding or removing electrons. Since electrons have mass ( $m_e = 9.1 \times 10^{-31}$ kg), any change in the number of electrons changes the body's total mass.

The sign of the charge tells us what happened:

Positive charge: electrons were removed → mass decreases
Negative charge: electrons were added → mass increases

The magnitude of charge tells us how many electrons moved, since each electron carries charge $e = 1.6 \times 10^{-19}$ C.

Finding the mass change

Determine the number of electrons involved

The charge acquired is $Q = 8.0 \times 10^{-12}$ C (positive). The number of electrons that must have been removed is:

n = \frac{Q}{e} = \frac{8.0 \times 10^{-12}}{1.6 \times 10^{-19}}

n = 5.0 \times 10^{7} \text{ electrons}

Calculate the total mass of these electrons

Each electron has mass $m_e = 9.1 \times 10^{-31}$ kg, so the total mass lost is:

\Delta m = n \times m_e = 5.0 \times 10^{7} \times 9.1 \times 10^{-31}

\Delta m = 45.5 \times 10^{-24} = 4.55 \times 10^{-23} \text{ kg}

Determine the direction of change

Since the body acquired positive charge, electrons left the body. Therefore, the mass decreases by $4.55 \times 10^{-23}$ kg.

Watch out

A common mistake is to forget that positive charge means electron loss, not gain. The body doesn't gain positive particles; it loses negative ones.

Tip

Quick check: the mass change in charging problems is always tiny (order $10^{-23}$ kg or smaller for typical lab charges) because electron mass is so small. Options with larger mass changes are usually wrong.

✓Final answer

4.55 \times 10^{-23}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

CBSE 2026Set V11 markMCQ
$Q. Which one of the following charge cannot exist on a body?$
(a) $2\,e$
(b) $3\,e$
(c) $3.5\,e$
(d) $-4\,e$
›Reveal solutionSolution

(c) $3.5\,e$

✓Final answer
$3.5\,e$

Electric charge is quantised: any charge on a body must be an integral multiple of the elementary charge $e$ , i.e. $q = n e$ with $n$ an integer ( $\pm 1, \pm 2, \dots$ ). Charges $2e$ , $3e$ and $-4e$ are integral multiples of $e$ and are allowed. $3.5\,e$ is a fractional (non-integral) multiple of $e$ , so it cannot exist on a body.
CBSE 2026Set ANNUAL1 markMCQ
$Q. Number of electrons in one coulomb charge are -$
(i) $6.25\times10^{18}$
(ii) $6.25\times10^{12}$
(iii) $6.25\times10^{10}$
(iv) $6.25\times10^{14}$
›Reveal solutionSolution

1 C of charge contains $1/e$ electrons.

The charge on one electron is $e=1.6\times10^{-19}$ C. The number of electrons whose total charge is 1 coulomb is $n=\dfrac{1}{e}=\dfrac{1}{1.6\times10^{-19}}=6.25\times10^{18}$ .

✓Final answer
$6.25\times10^{18}$

Note

The original paper's Hindi line asks for the number of electrons in one microcoulomb (1 µC), while the English line and the printed options are for one coulomb (1 C). This answer follows the English wording and the given options.
CBSE 2026Set DS1 markMCQ
$1 \times 10^{-7}$
i) $10^{7}$
ii) $1.6 \times 10^{19}$
iii) $6.25 \times 10^{11}$
iv) $9 \times 10^{12}$
›Reveal solutionSolution

Charge is quantised, so $n = Q/e = 6.25\times10^{11}$ electrons.

Concept. Charge exists only in whole multiples of the electronic charge $e = 1.6\times10^{-19}$ C (quantisation of charge). If a metal piece loses a charge $Q$ , the number of electrons removed is $n = \dfrac{Q}{e}$ .

Calculation.

$n = \frac{1\times10^{-7}}{1.6\times10^{-19}} = 0.625\times10^{12} = 6.25\times10^{11}.$

✓Final answer
$6.25\times10^{11}$
CBSE 2026Set ANNUAL1 markMCQ
$Q. An object has a negative charge of 1 coulomb. The number of excess electrons on it is$
(a) 6.25 x 10^-18
(b) 1.6 x 10^19
(c) 1.6 x 10^-19
(d) 6.25 x 10^18
›Reveal solutionSolution

Charge is quantised: any charge Q is made up of a whole number of electron charges e, so n = Q/e.

A negative charge means the object has more electrons than protons. Each electron carries a charge of magnitude e = 1.6 x 10^-19 C. If the object carries a total (excess) charge of magnitude Q, the number of excess electrons is

n = Q / e

Substituting Q = 1 C and e = 1.6 x 10^-19 C:

n = 1 / (1.6 x 10^-19) = 6.25 x 10^18

✓Final answer
$(d) 6.25 x 10^18 excess electrons.$
CBSE 2025Set 55/4/11 markMCQ
$8.0 \times 10^{-12}$
›Reveal solutionSolution
When a body gains positive charge, it loses electrons; the mass change equals the number of electrons lost times the electron mass. For $8.0 \times 10^{-12}$ C, the body decreases by $4.55 \times 10^{-23}$ kg.

Why charging changes mass

Charging a body means adding or removing electrons. Since electrons have mass ( $m_e = 9.1 \times 10^{-31}$ kg), any change in the number of electrons changes the body's total mass.

The sign of the charge tells us what happened:

Positive charge: electrons were removed → mass decreases

Negative charge: electrons were added → mass increases

The magnitude of charge tells us how many electrons moved, since each electron carries charge $e = 1.6 \times 10^{-19}$ C.

Finding the mass change

Determine the number of electrons involved

The charge acquired is $Q = 8.0 \times 10^{-12}$ C (positive). The number of electrons that must have been removed is:

$n = \frac{Q}{e} = \frac{8.0 \times 10^{-12}}{1.6 \times 10^{-19}}$ $n = 5.0 \times 10^{7} \text{ electrons}$

Calculate the total mass of these electrons

Each electron has mass $m_e = 9.1 \times 10^{-31}$ kg, so the total mass lost is:

$\Delta m = n \times m_e = 5.0 \times 10^{7} \times 9.1 \times 10^{-31}$ $\Delta m = 45.5 \times 10^{-24} = 4.55 \times 10^{-23} \text{ kg}$

Determine the direction of change

Since the body acquired positive charge, electrons left the body. Therefore, the mass decreases by $4.55 \times 10^{-23}$ kg.

Watch out

A common mistake is to forget that positive charge means electron loss, not gain. The body doesn't gain positive particles; it loses negative ones.

Tip

Quick check: the mass change in charging problems is always tiny (order $10^{-23}$ kg or smaller for typical lab charges) because electron mass is so small. Options with larger mass changes are usually wrong.

✓Final answer
$4.55 \times 10^{-23}$
CBSE 2025Set JS1 mark
$2.4 \times 10^{-18}$
›Reveal solutionSolution

A positive charge of $2.4\times10^{-18}$ C corresponds to a shortage of $n=q/e=15$ electrons.

Concept — quantisation of charge. Charge exists only in integer multiples of the electronic charge $e=1.6\times10^{-19}$ C: $q=ne$ . A positive conductor has lost electrons, so it has a deficit of electrons.

Solution.

$n=\frac{q}{e}=\frac{2.4\times10^{-18}}{1.6\times10^{-19}}=15.$

So the conductor is short of $15$ electrons.

✓Final answer
$15$
CBSE 2025Set A1 mark
$Q. Fill in the blank with appropriate word: One coulomb charge has ______ electrons.$
›Reveal solutionSolution

One coulomb of charge corresponds to about 6.25 × 10¹⁸ electrons.

Charge is quantised: total charge $q = ne$ , where $n$ is the number of electrons (or elementary charges) and $e = 1.6\times10^{-19}$ C is the magnitude of charge on a single electron. For $q = 1$ C:

$n = \dfrac{q}{e} = \dfrac{1}{1.6\times10^{-19}} = 6.25\times10^{18}$

So one coulomb of charge is carried by 6.25 × 10¹⁸ electrons.

✓Final answer
$6.25 \times 10¹⁸ electrons.$
CBSE 2025Set A1 mark
$Q. Write answer in one sentence: Write mathematical form of quantisation of electric charge.$
›Reveal solutionSolution

Quantisation of charge is expressed mathematically as q = ne.

Experiments (starting with Millikan's oil-drop experiment) show that electric charge is never continuous but always occurs in discrete, integral multiples of a smallest indivisible unit of charge, called the elementary charge $e = 1.6\times10^{-19}$ C (the magnitude of the charge on an electron or proton). This is expressed mathematically as:

$q = ne$

where n is any integer (positive, negative or zero) and e is the elementary charge. This means a body can only ever carry a charge that is a whole-number multiple of e; fractional charges (other than for quarks, which are never found free) do not occur in nature.

✓Final answer
$q = ne (n = integer, e = elementary charge = 1.6 \times 10⁻¹⁹ C).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. How many electrons will have a charge of one Coulomb?$
(a) 6.25 x 10^18
(b) 6.25 x 10^19
(c) 5.25 x 10^18
(d) 5.25 x 10^19
›Reveal solutionSolution

Charge is quantized in units of the electronic charge e = 1.6 x 10^-19 C, so the number of electrons needed to make up 1 C is n = Q/e.

Any charge Q is an integer multiple of the elementary charge: Q = n e.

Here Q = 1 C and e = 1.6 x 10^-19 C, so

n = Q/e = 1 / (1.6 x 10^-19) = 6.25 x 10^18

This huge number is why 1 coulomb is considered a very large amount of charge in practice - it takes about 6.25 billion billion electrons to make it up.

✓Final answer
$(a) 6.25 x 10^18 electrons.$
CBSE 2024Set A1 markMCQ
$Q. Number of electrons present in 8 coulomb negative charge is (A) 5 \times 10^19 (B) 2.5 \times 10^19 (C) 12.8 \times 10^19 (D) 1.6 \times 10^19$
›Reveal solutionSolution

n = Q/e = 8 ÷ (1.6×10⁻¹⁹) = 5×10¹⁹ electrons.

Charge is quantised: total charge $Q = n\,e$ , where $e = 1.6\times10^{-19}$ C.

$n = \frac{Q}{e} = \frac{8}{1.6\times10^{-19}} = 5\times10^{19}$

✓Final answer
$(A) 5 \times 10¹⁹.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The minimum amount of charge observed so far is$
(a) 1 C
(b) 4.8 x 10^-13 C
(c) 1.6 x 10^-19 C
(d) 1.6 x 10^19 C
›Reveal solutionSolution

Charge is quantised: every observable free charge is an integer multiple of the elementary charge e = 1.6 x 10^-19 C, the smallest charge ever measured on a free particle.

Millikan's oil-drop experiment established that electric charge does not take arbitrary values but always occurs as an integral multiple of a smallest unit, the electronic charge

$e = 1.6 \times 10^{-19}\ \text{C}$

This is the smallest amount of free, isolated charge ever detected in an experiment (quarks carry fractional charges of e, but they are never observed in isolation - they are always confined inside hadrons). So among the given options, 1 C is far too large to be a 'minimum', and 1.6 x 10^19 C is dimensionally the reciprocal (a distractor). The genuine minimum observed free charge is 1.6 x 10^-19 C.

✓Final answer
$(c) 1.6 x 10^-19 C.$
CBSE 2024Set ANNUAL1 mark
$q_1 + q_2 = 0$
›Reveal solutionSolution

$q_1+q_2=0$ means the two charges are equal in magnitude and opposite in sign, giving zero net charge for the pair — the defining condition of a system like an electric dipole.

If $q_1+q_2=0$ , then $q_2=-q_1$ : the two charges have exactly equal magnitude but opposite polarity (one is $+q$ , the other is $-q$ ). This does not mean there is no electric field around them — a pair of equal and opposite charges separated by some distance still produces a field (this is precisely the definition of an electric dipole, with dipole moment $p=q\times(\text{separation})$ , and it is only the algebraic (net) charge of the pair that vanishes, not the field. It also means that if this pair is enclosed inside a Gaussian surface, the total flux through that surface would be zero (Gauss's law), since only the enclosed net charge determines the flux.

✓Final answer
$It means the two charges are equal in magnitude and opposite in sign, so the system's net (total) charge is zero — the characteristic condition for an electric dipole.$

Why charging changes mass

Finding the mass change

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